# Finite morphism “with the right number of equations”

Suppose that $X$ is a scheme, say smooth, connected of finite type over a field of characteristic zero, $Y \to X$ is a smooth morphism of relative dimension $d$, and $f_1, \dots, f_d \in \mathcal{O}(Y)$ are $d$ global sections. Let $Z = Z(f_1, \dots, f_d) \subset Y$ be the closed (not necessarily reduced) subscheme defined by the global sections. Assume that the composite $Z \to X$ is finite.

What can be concluded in this situation? As far as I understand if $Z$ happened to be integral then it would be Cohen-Macaylay and flat over $X$. Is there something that can be said in greater generality? What if I assume that $X$ is the spectrum of a Henselian local ring, for example?

Related question: Are finite correspondances flat?

• Your assumption that $Z \to X$ is finite implies that $Z$ is lci (since $X$ and $Y$ are smooth). So, $Z$ is also Cohen--Macaulay, hence it is flat over $X$ without any further assumptions. – ulrich Sep 6 '16 at 10:22
• Are you saying that the $f_i$ automatically form a regular sequence? Do you have a reference? I am afraid my ignorance of geometry is showing :/. – Tom Bachmann Sep 6 '16 at 10:47
• Yes, the $f_i$ indeed form a regular sequence. See, for example, Theorem 2.12 (c) of the book "Cohen-Macaulay rings" by Bruns--Herzog (or any other commutative algebra book which discusses regular sequences). – ulrich Sep 7 '16 at 5:42
• Fantastic, thank you! If you post this as an answer I will accept it. – Tom Bachmann Sep 7 '16 at 9:17
• (For anyone else looking, it's theorem 2.1.2 (c) in the book cited.) – Tom Bachmann Sep 7 '16 at 9:53

Let z be a point of Z. Because f : Z --> X is finite, then OZ,z is an OX,f(z)-algebra of the same dimension. By assumption, OY,z is an OX,f(z)-algebra of relative dimension d.

By the theory of regular sequences (see the refernce in the comments), the sequence fi,z in OY,z, of stalks of the functions fi, is regular if the relative dimension of OY,z over OZ,z = OY,z/(fi,z) is equal to d. This is the case because of the assumptions. Therefor I conclude that the sequence is regular. This means that the embedding Z --> Y is a regular embedding. This implies that f : Z --> X is flat. Q.E.D.